Solving Trig Equations

[ilovemath3]

Hello once again, i had a few problems i was hoping i could check answseres with

1) cos 3x=0, i got the answeres to be pie/6 and 3pie/6

2) how woul di go about solving 5sinxtanx - 10tanx+3sinx-6=0??

3) also for 3+9+15+...+(6x-3)=(3x)Squard, after going through im supposed to get (3k+1)sq

i got it all the way down to (3k)sq + 6k - 2... but i dunno where to go from there

4) 1/ 1x2 + 1/2x3 + 1/3x4 +... + 1/(n(n+1) = n(n+1)

i got this down to K(k+2)+1 / K+1(k+2) and i dont know where to go, maybe im just not seeing some algebra?

once again any help is appreciated, thank you once again!!!!

 

[galactus]

2) how woul di go about solving 5sinxtanx - 10tanx+3sinx-6=0??

Factor it:

\(\displaystyle \L\\5sinxtanx-10tanx+3sinx-6=0\)

\(\displaystyle \L\\5tanx(sinx-2)+3(sinx-2)=0\)

\(\displaystyle \L\\(5tanx+3)(sinx-2)=0\)

 

[Mrspi]

Hello once again, i had a few problems i was hoping i could check answseres with


3) also for 3+9+15+...+(6x-3)=(3x)Squard, after going through im supposed to get (3k+1)sq

i got it all the way down to (3k)sq + 6k - 2... but i dunno where to go from there

I'm not sure what you mean by "I got it all the way down to....." since you showed no work.

I assume that you showed that
3 + 9 + 15 + .....+ (6x - 3) = 3x<SUP>2</SUP> NOT (3x)<SUP>2</SUP>!!!

is true when x = 1:
3 = 3(1)<SUP>2</SUP>
3 = 3

Assume that it is true for x = k, where k is a natural number and k > 1:
3 + 9 + 15 + ..... + (6k - 3) = 3k<SUP>2</SUP>

Now, let's add one more term, the (k + 1)th term, to both sides:
3 + 9 + 15 + .... + (6k - 3) + [6(k + 1) - 3] = 3k<SUP>2</SUP> + [6(k + 1) - 3]

From here on out, I'll work with the right side only. Simplify:
3k<SUP>2</SUP> + 6k + 6 - 3
3k<SUP>2</SUP> + 6k + 3

Now, factor:
3(k<SUP>2</SUP> + 2k + 1)
3(k + 1)<SUP>2</SUP>

So,
3 + 9 + 15 + ..... + [6(k + 1) - 3] = 3(k + 1)<SUP>2</SUP>

Since we've shown that the statement is true for 1, and whenever it is true for a natural number k > 1, it is also true for k + 1, we have proved that the statement is true for all natural numbers.

I hope this helps you.

 

[soroban]

Hello, ilovemath3!

No wonder you can't prove #3 and #4. . . There are typos!

3) Prove: \(\displaystyle \,3\,+\,9\,+\,15\,+\,\cdots\,+\,(6x\,-\,3)\;=\;\L3x^2\)

I assume you're using an Inductive proof.

Verify \(\displaystyle S(1):\;3\:=\:3\cdot1^2\) . . . True!

Assume \(\displaystyle S(k):\;3\,+\,9\,+\,15\,+\,\cdots\,+\,+\,(6k\,-\,3)\;=\;3k^2\)


Add \(\displaystyle 6k\,+\,3\) to both sides:
. . \(\displaystyle 3\,+\,9\,+\,15\,+\,\cdots\,+\,(6k\,-\,3)\,+\,(6k\,+\,3)\;=\;3k^2\,+\,6k\,+\,3\)

The right side is: \(\displaystyle \,3(k^2\,+\,2k\,+\,1)\;=\;3(k\,+\,1)^2\)

Our statement becomes: \(\displaystyle \,3\,+\,9\,+\,15\,+\,\cdots\,+\,(6k\,+\,3)\;=\;3(k\,+\,1)^2\)

. . and this is \(\displaystyle S(k+1)\) . . . The proof is complete!

Edit: Too fast for me, Mrspi!

4) Prove: \(\displaystyle \L\,\frac{1}{1\cdot2}\,+\,\frac{1}{2\cdot3}\,+\,\frac{1}{3\cdot4}\,+\,\cdots\,+\,\frac{1}{n(n+1)}\;=\;\L\frac{n}{n+1}\)

If we don't have to use Induction, there's a clever proof . . .

Note that the series is actually:
. . \(\displaystyle \:\left(\frac{1}{1}\,-\,\frac{1}{2}\right)\,+\,\left(\frac{1}{2}\,-\,\frac{1}{3}\right)\,+\,\left(\frac{1}{3}\,-\,\frac{1}{4}\right) \,+\,\cdots\,+\left(\frac{1}{n}\,-\,\frac{1}{n+1}\right)\;\) . . . a "telescoping" series

Nearly all the terms cancel out and we have: \(\displaystyle \L\,1\,-\,\frac{1}{n\,+\,1}\:=\:\frac{n}{n\,+\,1}\)

 

[galactus]

3) also for 3+9+15+...+(6x-3)=(3x)Squard, after going through im supposed to get (3k+1)sq

i got it all the way down to (3k)sq + 6k - 2... but i dunno where to go from there

\(\displaystyle \L\\3\underbrace{(1+3+5+....+(2n-1))}_{\text{sum of odd integers}}=3n^{2}\)

\(\displaystyle \L\\3\sum_{k=1}^{n}(2k-1)\Rightarrow{3\left(2\frac{n(n+1)}{2}-n\right)}=3n^{2}\)

Therefore, we have \(\displaystyle \L\\3n^{2}\)

 

[ilovemath3]

thank you all so much for the help, sorry i didn't put how i arrived at the answer, it was very late here... anywho

i followed along with all of your answers and got your answer... basically i found where i messed up

ok... so 2 and 3 i got

for #1... i have new answers, anyone confirm them? they would be:

theta = 5pie/5, 9pie/6 these are for pie/6 + (2pie/3)n
theta = 7pie/6 and 11pie/6 these are for 3pie/6 + (2pie/3)n

and #4 i just realized i had to factor k(k+2)+1 to get Ksq+2k+1 / k+1(K+2) which goes to (k+1)(k+1) / (k+1) (k+2) and the k+1 cancel leaving with (K+1)/(K+2) which is the asnwer i wanted